4

Let $f:X\to Y$ be a representable morphism of algebraic stacks over an algebraically closed field $k$. I wonder is there is a "formal criterion" for checking the properties formally smooth, unramified, étale. If $X,Y$ were schemes, we would phrase the criterion by means of the maps $$X(A)\to Y(A)\times_{Y(B)}X(B)$$ induced by small extensions $A\to B$ (of local Artin $k$-algebras with residue field $k$). Those are maps of sets when $X,Y$ are schemes. What if $X,Y$ are stacks, so that $X(A)$, say, is a groupoid?

Here is an attempt for formal étaleness:

We say $f:X\to Y$ is formally étale if the following holds. Let $j:\textrm{Spec }B\to \textrm{Spec }A$ be a small extension. Then, for all maps $g:\textrm{Spec }B\to X$ and $h:\textrm{Spec }A\to Y$ such that $fg\cong hj$ as objects of $Y(B)$, there exists a morphism $\textrm{Spec }A\to X$, unique up to isomorphism in $X(A)$, such that $h\cong f\alpha\in Y(B)$ and $g\cong \alpha j\in X(B)$.

Questions. Is this phrasing correct? Is it possible to translate it in terms of a functor $X(A)\to Y(A)\times_{Y(B)}X(B)$?

Thanks!

Brenin
  • 14,072
  • Is there a pre-existing definition? I would guess that your phrasing is not correct on the basis that it treats isomorphism as a property rather than a structure. Surely the correct definition would imply that $X (A) \to Y (A) \times_{Y (B)} X (B)$ is, say, an equivalence of groupoids when $X \to Y$ is formally étale. – Zhen Lin Jan 07 '16 at 07:55
  • I am not sure what you mean by "treating isomorphisms as a property". The pre-existing definition is: for every morphism $U\to Y$ from a scheme, the map $U\times_YX\to U$ is formally étale. – Brenin Jan 07 '16 at 08:01
  • Well, the point is that there can be more than one isomorphism between two objects, and you should specify which isomorphisms are being used. – Zhen Lin Jan 07 '16 at 09:22
  • I see your point. I am still unsure how one should phrase the criterion though, even including the isomorphisms as part of the data. Maybe someone knows a reference? – Brenin Jan 07 '16 at 19:05

0 Answers0